i am given {An} , {Bn} so
An->a
Bn-An->0
prove that Bn->a
??
Is this in $\displaystyle \mathbb{R}^n$ or in a generalized metric space? Ill assume its the former.
We know that by the definition of a convergent sequence that if $\displaystyle \xi_n\to\xi$ then $\displaystyle \forall{\varepsilon>0}$ there exists $\displaystyle N$ such that if $\displaystyle N\leqslant{n}$ then $\displaystyle d(\xi_n,\xi)<\varepsilon$
So we have that $\displaystyle \forall\varepsilon>0$ there exists a $\displaystyle N$ such that $\displaystyle N\leqslant{n}$ implies that $\displaystyle d\left(A_n,a\right)<\frac{\varepsilon}{2}$
And that $\displaystyle \forall\varepsilon>0$ there exists a $\displaystyle N_0$ such that if $\displaystyle N_0\leqslant{n}$ then $\displaystyle d\left(B_n-A_n,0\right)<\frac{\varepsilon}{2}$
Now it should be pretty apparent that since the metric on $\displaystyle \mathbb{R}^n$ is $\displaystyle \left|\bold{x}-\bold{y}\right|$ that $\displaystyle d\left(B_n-A_n,0\right)<\varepsilon\implies d\left(B_n,A_n\right)<\varepsilon$
So now consider
$\displaystyle d\left(B_n,a\right)<d\left(B_n,A_n\right)+d\left(A _n,a\right)<\frac{\varepsilon}{2}+\frac{\varepsilo n}{2}=\varepsilon$
For $\displaystyle n>\max\left\{N,N_0\right\}$
I'm sorry, I do not know what you mean? These are generic terms $\displaystyle \varepsilon$ and $\displaystyle \frac{\varepsilon}{2}$ are interchangable, they are both constants which you pick. You could have easily made both distances less than $\displaystyle \varepsilon$ resulting in [$\displaystyle d\left(B_n,a\right)<2\varepsilon$ it gives the same result